JAMB UTME - Physics - 1999

Ajụjụ 1 Ripọtì

The diagram above shows the prism arrangement in a

binocular

spectrometer

periscope

projector

Reflection Of Light At Plane And Curved Surfaces

Refraction Of Light Through Plane And Curved Surfaces

Lee Azịza

Ajụjụ 2 Ripọtì

At what frequency would a 10H inductor have a reactance of 2000Ω?

π/200Hz

π/100Hz

100/πHz

100πHz

Akọwa Nkọwa

Reactance is the opposition to the flow of alternating current caused by capacitance or inductance in a circuit. The reactance of an inductor is directly proportional to the inductance and the frequency of the current passing through it. The formula for the reactance of an inductor is X = 2πfL, where X is the reactance in ohms, f is the frequency in hertz, and L is the inductance in henries. To find the frequency at which a 10H inductor would have a reactance of 2000Ω, we can rearrange the formula to solve for f. X = 2πfL f = X/(2πL) Substituting the given values, we get: f = 2000/(2π × 10) f = 100/π Therefore, the frequency at which a 10H inductor would have a reactance of 2000Ω is 100/π Hz. Answer: 100/πHz

Gụọ akwụkwọ nkuzi na Elementary Modern Physics (JAMB)

Gụọ akwụkwọ nkuzi na Simple A.C Circuits (JAMB)

Elementary Modern Physics

Simple A.C Circuits

Lee Azịza

Ajụjụ 3 Ripọtì

The inside portion of part of a hollow metal sphere of diameter 20cm is polished. The portion will therefore form a

concave mirror of focal length 5cm

concave mirror of focal length 10cm

convex mirror of focal length 5cm

convex mirror of focal length 20cm

Akọwa Nkọwa

When a part of a hollow metal sphere is polished, it becomes reflective and can act as a mirror. The type of mirror it forms depends on the curvature of the surface. If the surface is curved inwards, it forms a concave mirror. If the surface is curved outwards, it forms a convex mirror. In this case, since the metal sphere is hollow and the inside portion is polished, the surface will be curved inwards, forming a concave mirror. To determine the focal length of this mirror, we need to know the radius of curvature of the mirror, which is half of the diameter of the original sphere. Thus, the radius of curvature is 10 cm. The focal length of a concave mirror is half of the radius of curvature, so the focal length of this mirror is 5 cm. Therefore, the answer is: concave mirror of focal length 5cm.

Lee Azịza

Ajụjụ 4 Ripọtì

in a series R-L-C circuit at resonance, the voltages across the resistor and the inductor are 30V and 40V respectively, what is the voltage across the capacitor?

30V

40V

50V

70V

Akọwa Nkọwa

In a series R-L-C circuit at resonance, the impedance of the circuit becomes purely resistive, and the total voltage of the circuit is divided between the resistor, inductor, and capacitor. At resonance, the voltage across the inductor is equal to the voltage across the capacitor. In this circuit, the voltage across the resistor is given as 30V, and the voltage across the inductor is given as 40V. Therefore, the total voltage across the circuit is 70V (30V + 40V). Since the voltage across the capacitor is equal to the voltage across the inductor at resonance, the voltage across the capacitor in this circuit is also 40V. Therefore, the answer is: 40V.

Lee Azịza

Ajụjụ 5 Ripọtì

The inner diameter of a test tube can be measured accurately using a

micrometer screw guage

pair of dividers

metre rule

pair of vernier calipers

Akọwa Nkọwa

To measure the inner diameter of a test tube accurately, a pair of vernier calipers should be used. A vernier caliper is a precision measuring tool that is used to take measurements with a high degree of accuracy. It consists of two jaws, one fixed and one movable, that can be adjusted using a fine screw adjustment mechanism. The scale on the caliper is read using a vernier scale, which allows for precise measurements to be taken. A micrometer screw gauge can also be used for very accurate measurements, but for measuring the inner diameter of a test tube, a pair of vernier calipers would be more appropriate. A metre rule or a pair of dividers would not be accurate enough for this measurement.

Gụọ akwụkwọ nkuzi na Measurements And Units (JAMB)

Measurements And Units

Lee Azịza

Ajụjụ 6 Ripọtì

The core of an efficient transformer should consist of laminated pieces of metal in order to

increase the heat produced by increasing the eddy current

increase the heat produced by reducing the eddy current

reduce the heat produced by increasing the eddy current

reduce the heat produced by reducing the eddy current

Akọwa Nkọwa

The core of an efficient transformer should consist of laminated pieces of metal in order to reduce the heat produced by reducing the eddy current. When a magnetic field changes in a conductor, eddy currents are produced. These eddy currents can cause heating in the core of a transformer and waste energy. By laminating the metal core, the electrical conductivity is reduced in the direction perpendicular to the lamination, which in turn reduces the eddy currents and thus the heat produced. This allows the transformer to operate more efficiently with less energy loss in the form of heat.

Gụọ akwụkwọ nkuzi na Properties Of Waves: Reflection, Refraction, Diffraction (WAEC)

Gụọ akwụkwọ nkuzi na Simple A.C Circuits (JAMB)

Properties Of Waves: Reflection, Refraction, Diffraction

Simple A.C Circuits

Lee Azịza

Ajụjụ 7 Ripọtì

Steel is more suitable for permanent magnet than iron because the former

is easily demagnetized by shaing vigorously

is an alloy of many metals

is easily magnetized by alternating current through one cycle

retains magnetism more than iron

Akọwa Nkọwa

The passage explains why steel is a better material for a permanent magnet than iron. It states that steel is more suitable for this purpose because it "retains magnetism more than iron." This means that once it is magnetized, it will maintain its magnetic properties for a longer period of time than iron would.

Gụọ akwụkwọ nkuzi na Magnets And Magnetic Fields (JAMB)

Magnets And Magnetic Fields

Lee Azịza

Ajụjụ 8 Ripọtì

Two bodies have masses in the ratio 3:1. They experience forces which impart to them acceleration in the ratio 2:9 respectively. Find the ratio of forces the masses experienced.

1 : 4

2 : 1

2 : 3

2 : 5

Akọwa Nkọwa

The ratio of forces acting on two bodies is equal to the ratio of their masses multiplied by the ratio of their accelerations. Let's denote the masses of the two bodies as 3m and m, where m is a constant of proportionality. We are given that the accelerations of the two bodies are in the ratio 2:9, which means that the acceleration of the first body is 2a and the acceleration of the second body is 9a, where a is a constant of proportionality. Using Newton's second law of motion, we know that force is equal to mass times acceleration. So, the force experienced by the first body is (3m)(2a) = 6ma, and the force experienced by the second body is (m)(9a) = 9ma. Therefore, the ratio of forces experienced by the two bodies is: 6ma : 9ma Simplifying this ratio by dividing both terms by 3m gives: 2a : 3a Which further simplifies to: 2:3 Therefore, the ratio of forces experienced by the two bodies is 2:3. So, the correct option is "2 : 3".

Gụọ akwụkwọ nkuzi na Temperature And Its Measurement (JAMB)

Temperature And Its Measurement

Lee Azịza

Ajụjụ 9 Ripọtì

What is the speed of a particle of mass 10-27kg whose wavelength is 10-8m.[h = 6.63 x 10-34Js]

6.63ms-1

66.30ms-1

663.00ms-1

6630.00ms-1

Akọwa Nkọwa

Energy of a particle is given by
E = hf = hV/λ (since V = fxλ)
But in terms of mass M, E = ∆MV2
∴ ∆MV2 = hV/λ
∆MV = h/λ
∴ V = h/λ∆M
V = 6.63 x 10-34/(10-27 x 10-8)
V = 6.63 x 10-34/10-35
V = 6.63 x 10-34+35
V = 6.63 x 10
V = 66.3ms-1

Gụọ akwụkwọ nkuzi na Electric Field (WAEC)

Electric Field

Lee Azịza

Ajụjụ 10 Ripọtì

A piece of substance of specific head capacity 450JKg-1K-1 falls through a vertical distance of 20m from rest. Calculate the rise in temperature of the substance on hitting the ground when all its energies are converted into heat. [g = 10ms-2]

2/9°C

4/9°C

9/4°C

9/2°C

Akọwa Nkọwa

Energy at the height = mgh
This energy is converted to heat energy - mcΔt
∴ From the principle of energy conversion,
mcΔt = mgh
m x 450 x Δt = m x 10 x 20
∴ Δt = m x 10 x 20/(m x 450)
= 200/450
= 4/9°C

Gụọ akwụkwọ nkuzi na Heat Energy (WAEC)

Heat Energy

Lee Azịza

Ajụjụ 11 Ripọtì

The diagram above shows a magnet X with its south pole moved along a soft - iron bar PQ in the direction shown. After some time, the poles at P and Q respectively are

North - North

North - South

South - North

South - South

Akọwa Nkọwa

When a magnet is moved along a soft-iron bar, the bar becomes temporarily magnetized. In this case, the south pole of the magnet X is moved towards point P of the bar PQ. As a result, the end of the bar near P becomes a north pole while the end near Q becomes a south pole. Therefore, the poles at P and Q respectively are North and South. So, the answer is "North - South".

Gụọ akwụkwọ nkuzi na Magnets And Magnetic Fields (JAMB)

Gụọ akwụkwọ nkuzi na Gravitational Field (WAEC)

Magnets And Magnetic Fields

Gravitational Field

Lee Azịza

Ajụjụ 12 Ripọtì

The figure above represents a frictionless pulley system in which a weight W is in equilibrium with a weight of 40N. Find the value of W.

13.3N

20.0N

40.0N

80.0N

Akọwa Nkọwa

In a frictionless pulley system, the tension in the string is constant throughout. Since the weight of 40N is being balanced by weight W, the tension in the string must be equal to 40N. Therefore, we have: Tension = 40N Also, we know that the weight W is balanced by the tension in the string. Therefore: W = Tension = 40N So the value of W is 40N. Answer: 40.0N

Gụọ akwụkwọ nkuzi na Scalars And Vectors (JAMB)

Gụọ akwụkwọ nkuzi na Equilibrium Of Forces (WAEC)

Scalars And Vectors

Equilibrium Of Forces

Lee Azịza

Ajụjụ 13 Ripọtì

2311Na + Proton → pqX + alpha particle

What are the values of p and q respectively in the equation above?

10 and 20

12 and 24

20 and 10

24 and 12

Akọwa Nkọwa

2311Na + Proton → pqX + alpha particle
Represent;
2311Na + 11P → pqX + 42He

Thus from the above equations, p = 20 and q = 10

Gụọ akwụkwọ nkuzi na Structure Of The Atom (Nigeria Only) (WAEC)

Structure Of The Atom (Nigeria Only)

Lee Azịza

Ajụjụ 14 Ripọtì

if ∆x is the uncertainty in the measurement of the position of a particle along the x-axis and ∆Px is the uncertainty in the measurement of the linear momentum along the x-axis, then the uncertainty principle relation is given as

∆x ∆Px ≥ h

∆x ∆Px = 0

∆x ∆Px< h

∆x ∆Px = ∞

Akọwa Nkọwa

The uncertainty principle is a fundamental principle of quantum mechanics that states that it is impossible to simultaneously determine the exact position and momentum of a particle. The uncertainty principle relation is given by the equation: ∆x ∆P ≥ h, where ∆x is the uncertainty in the measurement of the position of a particle along the x-axis, and ∆P is the uncertainty in the measurement of the linear momentum along the x-axis. The symbol h represents Planck's constant, which is a fundamental constant of nature. Therefore, the option that represents the uncertainty principle relation correctly is: ∆x ∆P ≥ h.

Gụọ akwụkwọ nkuzi na Position, Distance And Displacement (WAEC)

Position, Distance And Displacement

Lee Azịza

Ajụjụ 15 Ripọtì

Particles of mass 10-2kg is fixed to the tip of a fan blade which rotates with angular velocity of 100rad-1. If the radius of the blade is 0.2m, the centripetal force is

2 N

20 N

200 N

400 N

Akọwa Nkọwa

The centripetal force is the force that keeps an object moving in a circular path. It is given by the formula F = mrω^2, where F is the centripetal force, m is the mass of the object, r is the radius of the circular path, and ω is the angular velocity of the object. In this case, the mass of the particles is 10^-2 kg, the radius of the blade is 0.2m, and the angular velocity of the blade is 100 rad^-1. Plugging these values into the formula, we get: F = (10^-2 kg)(0.2 m)(100 rad^-1)^2 = 20 N Therefore, the centripetal force is 20 N.

Gụọ akwụkwọ nkuzi na Motion (JAMB)

Gụọ akwụkwọ nkuzi na Temperature And Its Measurement (JAMB)

Motion

Temperature And Its Measurement

Lee Azịza

Ajụjụ 16 Ripọtì

Gamma rays are produced when

high velocity electrons are abruptly stopped in metals

energy changes occur within the nucleus of atoms

energy changes occur within the electronic structure of atoms

electrons are deflected in very strong magnetic fields

Akọwa Nkọwa

Gamma rays are produced when energy changes occur within the nucleus of atoms. Gamma rays are electromagnetic radiation of high frequency and energy, and they are produced by nuclear reactions, such as radioactive decay or nuclear fusion. When there is a change in the nucleus of an atom, for example, when a nucleus emits a particle or undergoes a rearrangement of its protons and neutrons, it can release gamma rays. These gamma rays are highly energetic and can penetrate through many materials, making them useful in a variety of fields, including medicine and industry.

Gụọ akwụkwọ nkuzi na Structure Of The Atom (Nigeria Only) (WAEC)

Gụọ akwụkwọ nkuzi na Electromagnetic Field (Part 2) (WAEC)

Structure Of The Atom (Nigeria Only)

Electromagnetic Field (Part 2)

Lee Azịza

Ajụjụ 17 Ripọtì

Four cells each of e.m.f 1.5V and internal resistance of 4Ω are connected in parallel. What is the effective e.m.f. and internal resistance of the combination?

6.0V, 16Ω

6.0V, 1Ω

1.5V, 4Ω

1.5V, 1Ω

Akọwa Nkọwa

Since all the cells are connected in parallel, the effective e.m.f. is just that of one single cell. Which means e.m.f. = 1.5V

However, the effective internal resistance adds as parallel arrangement given as
1/r = 1/r1 + 1/r2 + 1/r3 + 1/r4
= 1/4 + 1/4 + 1/4 + 1/4
= 1Ω
∴ r = 1Ω

Answer = 1.5V, 1Ω

Gụọ akwụkwọ nkuzi na Electric Cells (JAMB)

Electric Cells

Lee Azịza

Ajụjụ 18 Ripọtì

A galvanometer has a resistance of 5Ω. By using a shunt wire of resistance 0.05Ω, the galvanometer could be converted to an ammeter capable of reading 2Amp. What is the current through the galvanometer?

2mA

10mA

20mA

25mA

Akọwa Nkọwa

Shunt resistance is usually given as
R = igrg/(I - ig)
Where R = 0.05Ω
rg = 5Ω
I = 2M
∴ ig = IR/(rg + R)
= (2 x 0.05)/(5 + 0.05)
= 0.10/5.05
= 0.0198A
= 19.8mA
= 20mA

Gụọ akwụkwọ nkuzi na Electric Cells (JAMB)

Electric Cells

Lee Azịza

Ajụjụ 19 Ripọtì

If the stress on a wire is 107NM-2 and the wire is stretched from its original length of 10.00 cm to 10.05 cm. The young's modulus of the wire is

5 . 0 x 104 Nm-2

5 . 0 x 105 Nm-2

2 . 0 x 108 Nm-2

2 . 0 x 109 Nm-2

Akọwa Nkọwa

Y = stress/strain, but stress = 107Nm-2 (given).

Strain = e/l0. = (10.05 - 10.00)/10.00 = 0.005

Thus Y = 107/0.005 = 2 . 0 x 109Nm-2

Gụọ akwụkwọ nkuzi na Elasticity (JAMB)

Elasticity

Lee Azịza

Ajụjụ 20 Ripọtì

If a sound wave goes from a cold air region to a hot air region, its wavelength will

increase

decrease

decrease then increase

remain constant

Akọwa Nkọwa

If a sound wave goes from a cold air region to a hot air region (which will be less dense), the velocity decreases, and since the frequency is always constant, the wave length decreases with the velocity

Gụọ akwụkwọ nkuzi na Current Electricity (JAMB)

Current Electricity

Lee Azịza

Ajụjụ 21 Ripọtì

When the brakes in a car are applied, the frictional force force on the tyres is

is a disadvantage because it is in the direction of the motion of the car

is a disadvantage because it is in the opposite direction of the motion of the car

is an advantage because it is in the direction of the motion of the car

is an advantage because it is in the opposite direction of the motion of the car

Friction

Lee Azịza

Ajụjụ 22 Ripọtì

In the J-tube above, Y and X are on the same horizontal level and 30cm3 of air is trapped above Y when the atmospheric pressure is 75cm Hg, calculate the volume of air trapped above Y when 15cm Hg is now poured into the limb above X

15 cm3

25 cm3

35 cm3

45 cm3

Akọwa Nkọwa

The J-tube is a device used to measure the pressure of gases. In this case, we have 30cm3 of air trapped above Y when the atmospheric pressure is 75cm Hg. When 15cm Hg is added to the limb above X, the pressure on both sides of the J-tube will be equalized, and the volume of air trapped above Y will change. To find the new volume of air trapped above Y, we can use Boyle's law, which states that the pressure of a gas is inversely proportional to its volume when the temperature is constant. Therefore, we can set up the equation: P1 x V1 = P2 x V2 where P1 and V1 are the initial pressure and volume of the air trapped above Y, and P2 and V2 are the final pressure and volume of the air trapped above Y. Plugging in the values given, we get: 75 cm Hg x 30 cm3 = (75 cm Hg + 15 cm Hg) x V2 Simplifying the equation: V2 = (75 cm Hg x 30 cm3) / (75 cm Hg + 15 cm Hg) V2 = (75 cm Hg x 30 cm3) / 90 cm Hg V2 = 25 cm3 Therefore, the volume of air trapped above Y when 15cm Hg is poured into the limb above X is 25 cm3. Answer: 25 cm3.

Gụọ akwụkwọ nkuzi na Pressure (JAMB)

Pressure

Lee Azịza

Ajụjụ 23 Ripọtì

A ball of mass 0.1kg is thrown vertically upwards with a speed of 10ms-1 from the top of a tower 10m high. Neglecting air resistance, its total energy just before hitting the ground is(take g = 10ms-2)

5 J

10 J

15 J

20 J

Akọwa Nkọwa

Height h = 1/2 (U2/g) = 102/20 = 5m

Energy before hitting ground is = P.E at the height of 15m.

P.E = mgh = 0.1 x 10 x 15 = 15 J

Gụọ akwụkwọ nkuzi na Work, Energy And Power (WAEC)

Work, Energy And Power

Lee Azịza

Ajụjụ 24 Ripọtì

The velocities of light in air and glass are 3.0 x 108ms-1 and 2.0 x 108ms-1 respectively. If the angle of refraction is 30°, the sine of the angle of incidence is

0.33

0.50

0.67

0.75

Akọwa Nkọwa

The sine of the angle of incidence can be found by applying Snell's law, which states that the ratio of the sines of the angles of incidence and refraction is equal to the ratio of the velocities of light in the two media. So, we have: sin(angle of incidence) / sin(30°) = velocity of light in air / velocity of light in glass Substituting the given values, we get: sin(angle of incidence) / 0.5 = (3.0 x 10^8 m/s) / (2.0 x 10^8 m/s) Solving for sin(angle of incidence), we get: sin(angle of incidence) = 0.5 x (3.0 / 2.0) = 0.75 Therefore, the sine of the angle of incidence is 0.75. So, the answer is 0.75.

Gụọ akwụkwọ nkuzi na Electromagnetic Waves: Types Of Radiation In Electromagnetic Spectrum (WAEC)

Electromagnetic Waves: Types Of Radiation In Electromagnetic Spectrum

Lee Azịza

Ajụjụ 25 Ripọtì

Which of the following electromagnetic waves is least energetic?

Infra-red rays

X-rays

Ultra-violet rays

Gamma rays

Akọwa Nkọwa

The least energetic electromagnetic wave is the one with the lowest frequency and wavelength. In the given options, the electromagnetic waves are arranged in increasing order of frequency and decreasing order of wavelength. Therefore, the electromagnetic wave that is least energetic is Infrared rays.

Gụọ akwụkwọ nkuzi na Electromagnetic Waves: Types Of Radiation In Electromagnetic Spectrum (WAEC)

Gụọ akwụkwọ nkuzi na Electromagnetic Field (Part 2) (WAEC)

Electromagnetic Waves: Types Of Radiation In Electromagnetic Spectrum

Electromagnetic Field (Part 2)

Lee Azịza

Ajụjụ 26 Ripọtì

A car of mass 800kg attains a speed of 25m/s in 20 secs. The power developed in the engine is

1 . 25 x 104 W

2 . 50 x 104 W

1 . 25 x 106W

2 . 50 x 106 W

Akọwa Nkọwa

Power = Force x

\frac{d i s t a n c e}{t i m e}

=

\frac{m (v - u)}{t} \times v

$\frac{800 (25 - 0)}{20} \times 25$ = 2.5 x 10 4 watts

Gụọ akwụkwọ nkuzi na Work, Energy And Power (WAEC)

Work, Energy And Power

Lee Azịza

Ajụjụ 27 Ripọtì

The diagram above shows the velocity-time graph of a vehicle. Its acceleration and retardation respectively are

8.0 ms-2, 4,0 ms-2

4.0 ms-2, 8.0 ms-2

4.0 ms-2, 2.0 ms-2

2.0 ms-2, 4.0 ms-2

Akọwa Nkọwa

Acceleration = slope = AE/OE = 80/20 = 4m/s2
∴ acceleration and retardation = 4m/s2; 2m/s2

Gụọ akwụkwọ nkuzi na Rectilinear Acceleration (WAEC)

Rectilinear Acceleration

Lee Azịza

Ajụjụ 28 Ripọtì

The diagram above shows a solid figure with base PQ and center of gravity G on a inclined plane. Which of the following statements is correct

The solid will fall over if the vertical line through G lies outside the base

The solid will fall over if the vertical line through lies inside the base

The solid will not fall over if the vertical line through G lies outside the base

The solid can never fall over

Akọwa Nkọwa

The stability of a solid placed on an inclined plane depends on the position of its center of gravity with respect to the base. The base of the solid is the polygon formed by the intersection of the solid and the plane on which it rests. The center of gravity of the solid is the point where its weight can be considered to act. If the vertical line through the center of gravity G falls within the base PQ, the solid will be in a stable equilibrium, meaning it will not fall over. However, if the vertical line through G falls outside the base, the solid will be in an unstable equilibrium, meaning it will topple over. Therefore, option (A) is the correct statement: "The solid will fall over if the vertical line through G lies outside the base."

Gụọ akwụkwọ nkuzi na Equilibrium Of Forces (WAEC)

Equilibrium Of Forces

Lee Azịza

Ajụjụ 29 Ripọtì

In the circuit above, The potential across each capacitor is 100V. The total energy stored in the two capacitors is

3.0 x 104J

3.0 x 102J

2.5 x 10-2J

6.0 x 10-3J

Akọwa Nkọwa

For capacitor in parallel we have C=C1+C2+C3. so we C=2+3 =5UF. E=1/2CVsquare

Gụọ akwụkwọ nkuzi na Capacitors (JAMB)

Capacitors

Lee Azịza

Ajụjụ 30 Ripọtì

A gas at a volume of V0 in a container at pressure P0 is compressed to one-fifth of its volume. What will be its pressure if the magnitude of its original temperature T is constant?

P0/5

4P0/5

P0

5P0

Akọwa Nkọwa

At constant T, P1 V1 = P2 V2.

Let P1 = P0 and V1 = V0, then solve accordingly.

Gụọ akwụkwọ nkuzi na Quantity Of Heat (JAMB)

Quantity Of Heat

Lee Azịza

Ajụjụ 31 Ripọtì

In the diagram above, if the internal resistance of the cell is zero, the ration of the powers P1 and P2 dissipated by R1 and R2

R2/R1

R1/R2

R1+R2/R1

R1+R2/R2

Akọwa Nkọwa

If the internal resistance of the cell is zero, then the potential difference across the cell is equal to the potential difference across R1 and R2 combined. Therefore, the current passing through R1 and R2 will be in the inverse ratio of their resistances. The power dissipated in a resistor is proportional to the square of the current passing through it. So, the ratio of the powers P1 and P2 dissipated by R1 and R2 will be equal to the square of the inverse ratio of their resistances. Thus, the answer is: R2/R1.

Gụọ akwụkwọ nkuzi na Current Electricity (JAMB)

Current Electricity

Lee Azịza

Ajụjụ 32 Ripọtì

In the Rutherford scattering experiment, a beam of alpha particles was fired at a thin gold film with some of the particles being considerably deflected. This shows that

a gold nucleus contains protons, neutrons and electrons uniformly distributed in a tiny volume

the gold nucleus is positively charged and is concentrated in a tiny volume

the gold nucleus emitted alpha particles

the gold nucleus is concentrated in a tiny volume and contains alpha particles

Akọwa Nkọwa

In the Rutherford scattering experiment, a beam of alpha particles was fired at a thin gold film with some of the particles being considerably deflected. This shows that the gold nucleus is positively charged and is concentrated in a tiny volume. Rutherford's experiment provided evidence that the atom is mostly empty space, with a very small and dense positively charged nucleus at the center. The positive charge of the nucleus causes alpha particles to be deflected when they get close to it. The deflection observed in the experiment was much greater than expected if the positive charge were distributed uniformly throughout the gold foil, indicating that the charge is concentrated in a tiny volume at the center of the atom.

Gụọ akwụkwọ nkuzi na Structure Of The Atom (Nigeria Only) (WAEC)

Structure Of The Atom (Nigeria Only)

Lee Azịza

Ajụjụ 33 Ripọtì

The diagram above shows a lift pump with valves at P and Q. During a downward stroke of the piston

both valves are open

P is open while Q is closed

P is closed while Q is open

Both valves are closed

Akọwa Nkọwa

In a lift pump, the piston moves up and down inside a cylinder, which is connected to two valves at the bottom. When the piston is moving downwards, it creates a low-pressure region inside the cylinder which causes the valve at the bottom, marked as P, to open and the valve at the top, marked as Q, to close. This allows water to enter the cylinder through the valve P. During the upward stroke of the piston, valve P closes and valve Q opens, allowing the water to be lifted up and out of the cylinder. Therefore, during a downward stroke of the piston, valve P is open while valve Q is closed. The answer is P is open while Q is closed.

Lee Azịza

Ajụjụ 34 Ripọtì

Calculate the mass of ice that would melt when 2kg of copper is quickly transferred from boiling water to a block of ice without heat loss;

Specific heat capacity of copper = 400JKg-1K-1

Latent heat of fusion of ice = 3.3 x 105JKg-1

8/33kg

33/80kg

80/33kg

33/8kg

Akọwa Nkọwa

The problem involves the transfer of heat from hot copper to a block of ice resulting in the melting of some ice. The heat lost by the copper is equal to the heat gained by the ice that melts. Therefore, we can use the equation Q = mL, where Q is the heat transferred, m is the mass of ice melted, and L is the latent heat of fusion of ice. First, we need to find the heat lost by the copper. This can be calculated using the equation Q = mcΔT, where m is the mass of copper, c is its specific heat capacity, and ΔT is the change in temperature. Since the copper is transferred from boiling water, its initial temperature is 100°C. When it comes in contact with ice, its temperature drops to 0°C. Therefore, ΔT = 100 - 0 = 100°C. Substituting the given values, we get: Q = (2 kg) x (400 J/kg.K) x (100°C) = 80000 J Now, we can use the equation Q = mL to find the mass of ice melted. Substituting the given values, we get: 80000 J = m x (3.3 x 10^5 J/kg) m = 80000 J ÷ (3.3 x 10^5 J/kg) = 0.2424 kg or 242.4 g Therefore, the mass of ice that would melt is 242.4 g, which is equivalent to 8/33 kg (rounded to three significant figures). Answer: 8/33kg

Gụọ akwụkwọ nkuzi na Heat Energy (WAEC)

Heat Energy

Lee Azịza

Ajụjụ 35 Ripọtì

If ∆x is the uncertainty in the measurement of the position of a particle along the x-axis and ∆Pa is the uncertainty in the measurement of the linear momentum along the x-axis, then the uncertainty principle relation is given as

∆x ∆Px ≥ h

∆x ∆Px = 0

∆x ∆Px< h

∆x ∆Px = ∞

Akọwa Nkọwa

The uncertainty principle is a fundamental principle of quantum mechanics that states that it is impossible to simultaneously determine certain pairs of physical properties of a particle, such as its position and momentum, with perfect accuracy. The uncertainty principle relation between the uncertainty in the measurement of the position (∆x) and the uncertainty in the measurement of the linear momentum (∆P) along the x-axis is given by the equation: ∆x ∆P ≥ h where h is Planck's constant. This means that the product of the uncertainties in the measurement of position and momentum cannot be smaller than h. Therefore, option A is the correct answer.

Gụọ akwụkwọ nkuzi na Structure Of The Atom (Nigeria Only) (WAEC)

Structure Of The Atom (Nigeria Only)

Lee Azịza

Ajụjụ 36 Ripọtì

The equation of a wave travelling along the positive x-direction is given by;

Y = 0.25 x 10-3 sin (500t - 0.025x).

Determine the angular frequency of the wave motion.

0.25 x 10-3rad s-1

0.25 x 10-1rad s-1

5.00 x 102rad s-1

2.50 x 103rad s-1

Akọwa Nkọwa

The equation of the wave is given as Y = 0.25 x 10^-3 sin (500t - 0.025x). To determine the angular frequency of the wave motion, we need to find the coefficient of 't' in the argument of the sine function, which is 500. The angular frequency is then given by 2π times the frequency, where the frequency is 500/2π. Therefore, the angular frequency of the wave is 5.00 x 10² rad s^-1. So, the correct option is: - 5.00 x 10² rad s^-1

Gụọ akwụkwọ nkuzi na Production And Propagation Of Waves (WAEC)

Production And Propagation Of Waves

Lee Azịza

Ajụjụ 37 Ripọtì

The diagram above represents a transverse electromagnetic wave travelling with speed 3.0 x 108ms-1. What is the frequency of the wave.

3.0 x 107Hz

9.0 x 107Hz

1.0 x 109Hz

3.0 x 109Hz

Akọwa Nkọwa

The frequency f = V/λ,
Where λ = wavelength
From the above, the distance OA = one wavelength; AB = one wave length and BC = one wavelength.
=> OC = 3 wavelength = 3λ
∴3λ = 0.3M
λ = 0.3/3
= 0.1m
∴ f = V/λ = (3.0 x 108) / 0.1
= 3.0 x 109Hz

Gụọ akwụkwọ nkuzi na Electromagnetic Waves: Types Of Radiation In Electromagnetic Spectrum (WAEC)

Electromagnetic Waves: Types Of Radiation In Electromagnetic Spectrum

Lee Azịza

Ajụjụ 38 Ripọtì

A 3000W electric cooker is to be used on a 200V mains circuit. Which of the fuses below can be used safely with the cooker?

2A

5A

10A

20A

Akọwa Nkọwa

The maximum current drawn by the cooker is given by;
I = P/V
∴ I = 3000/200
I = 15A
Thus the most suitable fuse to be used is a 10A fuse

Gụọ akwụkwọ nkuzi na Electrical Energy And Power (JAMB)

Electrical Energy And Power

Lee Azịza

Ajụjụ 39 Ripọtì

a lead bullet of mass 0.05kg is fired with a velocity of 200ms-1 into a lead block of mass 0.95kg. Given that the lead block can move freely, the final kinetic energy after impact is

50 J

100 J

150 J

200 J

Akọwa Nkọwa

From principle of conservation of linear momentum,
(0.05 x 200) - (0.95 x 0) = (0.05 + 0.95) x V (since collision is inelastic).

Thus V = 10m/s.

K.E = 1/2 (0.05 + 0.95) x 102

K.E = 1/2 (1 x 100) = 50 J

Gụọ akwụkwọ nkuzi na Gravitational Field (JAMB)

Gụọ akwụkwọ nkuzi na Work, Energy And Power (WAEC)

Gravitational Field

Work, Energy And Power

Lee Azịza

Ajụjụ 40 Ripọtì

In fleming's right-hand rule, the thumb, the forefinger and the middle finger if held mutually at right angles represent respectively, the

motion, the field and the induced current

induced current, the motion and the field

field, the induced current and the motion

induced current, the field and the motion

Akọwa Nkọwa

Fleming's right-hand rule is a rule used to determine the direction of motion, current, and magnetic field in an electric generator or motor. According to the rule, if the thumb, forefinger, and middle finger of the right hand are held mutually at right angles, they represent respectively, the motion, the field, and the induced current. Therefore, the correct option is: "motion, the field and the induced current".

Gụọ akwụkwọ nkuzi na Electrostatics (JAMB)

Electrostatics

Lee Azịza

Ajụjụ 41 Ripọtì

An astronomical telescope is said to be in normal adjustment when the

eye is accommodated

focal length of objective lens is longer than that of eye piece

final image is at the near point of eye

final image is at infinity

Akọwa Nkọwa

An astronomical telescope is said to be in normal adjustment when the final image is at infinity. In an astronomical telescope, the objective lens forms a real image of the object at its focus which is farther from the objective lens. The eyepiece is then used as a magnifying glass to form a virtual image of the real image, which is closer to the eye. When the virtual image is formed at infinity, the eye is relaxed, and the eye lens is effectively at its focal length, allowing for maximum sharpness and clarity.

Lee Azịza

Ajụjụ 42 Ripọtì

In the circuit above, the ammeter reads a current of 5.0 A when R = 8? and reads 7.0A when R = 5?. The value of the unknown resistance X is

10.0?

7.5?

5.0?

2.5?

Akọwa Nkọwa

Gụọ akwụkwọ nkuzi na Capacitors (JAMB)

Gụọ akwụkwọ nkuzi na Simple A.C Circuits (JAMB)

Capacitors

Simple A.C Circuits

Lee Azịza

Ajụjụ 43 Ripọtì

The lowest note emitted by a stretched string has a frequency of 40Hz. How many overtones are there between 40hz and 180Hz?

4

3

2

1

Akọwa Nkọwa

The first overtone of a stretched string occurs at twice the frequency of the fundamental (lowest) frequency. Therefore, the first overtone of the string is at 80Hz, the second overtone is at 120Hz, and the third overtone is at 160Hz. Since the question asks for the number of overtones between 40Hz and 180Hz, we can count that there are three overtones within this range (80Hz, 120Hz, and 160Hz), so the answer is 3.

Gụọ akwụkwọ nkuzi na Production And Propagation Of Waves (WAEC)

Production And Propagation Of Waves

Lee Azịza

Ajụjụ 44 Ripọtì

The temperature gradient across a copper rod of thickness 0.02m, maintained at two temperature junctions of 20°C and 80°C respectively is

3.0 x 102Km-1

3.0 x 103Km-1

5.0 x 103Km-1

3.0 x 104Km-1

Akọwa Nkọwa

Temperature Gradient =

\frac{Δ temp.}{thickness}

= $\frac{80 - 20}{0.02}$

= $\frac{60}{0.02}$

= 3000Km-1

= 3.0 x 103Km-1

Gụọ akwụkwọ nkuzi na Temperature And Its Measurement (JAMB)

Temperature And Its Measurement

Lee Azịza

Ajụjụ 45 Ripọtì

I. A liquid boils when its saturated vapour pressure is equal to the external pressure

II. Dissolved substances in pure water lead to increase in the boiling point.

III. When the external pressure is increased, the boiling point increases.

IV. Dissolved substances in pure water decrease the boiling point.Which combination of the above are peculiarities of the boiling point of a liquid?

I, II and III

I, II, III and IV

I, II and IV

II, III and IV

Akọwa Nkọwa

The boiling point of a liquid is a physical property that describes the temperature at which a liquid changes to its vapor state. The peculiarities of boiling point of a liquid can be determined by considering the conditions that affect the boiling process. Statement I says that a liquid boils when its saturated vapor pressure is equal to the external pressure. This is a correct statement as it implies that a liquid will boil when the vapor pressure above the liquid becomes equal to the pressure surrounding the liquid. Statement II states that dissolved substances in pure water lead to an increase in the boiling point. This is also a correct statement. When a substance is dissolved in a liquid, it causes the boiling point of the liquid to increase. Statement III states that when the external pressure is increased, the boiling point increases. This is a correct statement as well. When external pressure is increased, the boiling point of a liquid also increases. Statement IV states that dissolved substances in pure water decrease the boiling point. This statement is incorrect as dissolved substances in pure water increase the boiling point of water. Therefore, the combination of statements that are peculiarities of the boiling point of a liquid are I, II, and III.

Gụọ akwụkwọ nkuzi na Quantity Of Heat (JAMB)

Quantity Of Heat

Lee Azịza

Ajụjụ 46 Ripọtì

In the series a.c circuit shown above, the p.d across the inductor is 8Vr.m.s and that across the resistor is 6Vr.m.s. The effective voltage is

2V

10V

14V

48V

Akọwa Nkọwa

To find the effective voltage in an AC circuit with both inductor and resistor, we need to use the concept of phasor diagram. The effective voltage is the hypotenuse of the right-angled triangle formed by the phasor diagram, where the vertical side represents the voltage across the inductor, and the horizontal side represents the voltage across the resistor. Using the given values, we can draw the phasor diagram and calculate the effective voltage as follows: - Let the voltage across the inductor be represented by a vertical phasor of 8V. - Let the voltage across the resistor be represented by a horizontal phasor of 6V. - The hypotenuse of the right-angled triangle formed by the two phasors represents the effective voltage. - Using the Pythagorean theorem, we can find the length of the hypotenuse: - hypotenuse = sqrt(vertical side^2 + horizontal side^2) - hypotenuse = sqrt(8^2 + 6^2) = sqrt(100) = 10V Therefore, the effective voltage in the circuit is 10V. So, the correct option is: - 10V

Lee Azịza

Ajụjụ 47 Ripọtì

A semiconductor diode is used in rectifying alternating current into direct current mainly because it

allows current to flow in either direction

is non-linear

offers a high input resistance

allows current to flow only in one direction

Akọwa Nkọwa

A semiconductor diode is used in rectifying alternating current into direct current mainly because it allows current to flow only in one direction. This is because a diode has a characteristic property of conducting current in one direction and blocking it in the opposite direction. When an AC voltage is applied across the diode, it allows the current to flow only during the positive half cycle and blocks it during the negative half cycle. As a result, the diode acts as a rectifier, converting AC voltage into DC voltage.

Gụọ akwụkwọ nkuzi na Simple A.C Circuits (JAMB)

Gụọ akwụkwọ nkuzi na Electrostatics (JAMB)

Simple A.C Circuits

Electrostatics

Lee Azịza

Ajụjụ 48 Ripọtì

A solid weigh 10 .00 N in air, 6 N when fully immersed in water and 7 . 0 N when fully immersed in a liquid X. Calculate the relative density of the liquid, X.

5/3

4/3

3/4

7/10

Akọwa Nkọwa

R.D = loss in weight in liquid X loss in weight in water.

R.D = (10 - 7)/(10 - 6) = 3/4

Gụọ akwụkwọ nkuzi na Temperature And Its Measurement (JAMB)

Temperature And Its Measurement

Lee Azịza

Ajụjụ 49 Ripọtì

When temperature of a liquid increases, its surface tension

Decreases

Increases

Remain Constant

Increases then decreases

Akọwa Nkọwa

When the temperature of a liquid increases, the average kinetic energy of the molecules in the liquid also increases. This causes the molecules to move more rapidly and break the intermolecular forces holding them together. This leads to a reduction in the attractive forces at the surface of the liquid and a decrease in the cohesive forces within the liquid, causing a decrease in surface tension. Therefore, the answer is: Decreases.

Gụọ akwụkwọ nkuzi na Temperature And Its Measurement (JAMB)

Temperature And Its Measurement

Lee Azịza

Ajụjụ 50 Ripọtì

A man stands 4m in front of a plane mirror. If the mirror is moved 1m towards the man, the distance between him and his new image is

3m

5m

6m

10m

Akọwa Nkọwa

The image of an object placed in front of a plane mirror is always as far behind the mirror as the object is in front of the mirror. Thus when the object is 4m in front of the mirror, the image will be 4m behind the mirror. But when the mirror is moved in towards the man, (that is the man is now 3m from the mirror), the image distance will again decrease by 1m from behind the mirror. Thus the new image position is 3m behind the mirror. Therefore, the distance between the man and his new image is 6m.

Gụọ akwụkwọ nkuzi na Motion (JAMB)

Gụọ akwụkwọ nkuzi na Liquids At Rest (JAMB)

Motion

Liquids At Rest

Lee Azịza






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